To solve the problem of finding the 2004th number in the list of all 7-digit numbers formed by the digits 1, 2, 3, 4, 5, 6, and 7 exactly once, and which are not divisible by 5, we can follow these steps:
### Step 1: Determine the total number of valid 7-digit numbers
Since the numbers cannot end with 5 (to ensure they are not divisible by 5), the valid digits for the unit place are 1, 2, 3, 4, 6, and 7. This gives us 6 choices for the last digit.
For the first 6 digits, we can use all 7 digits (1, 2, 3, 4, 5, 6, 7) except the digit chosen for the last place. Thus, the total number of valid arrangements can be calculated as follows:
- Choose a digit for the unit place (6 choices).
- Arrange the remaining 6 digits in the first 6 places (which can be done in \(6!\) ways).
So, the total number of valid 7-digit numbers is:
\[
6 \times 6! = 6 \times 720 = 4320
\]
### Step 2: Determine how many numbers start with each digit
Next, we need to find out how many numbers start with each digit.
1. **Starting with 1**: The unit place can be any of 2, 3, 4, 5, 6, 7 (6 choices). The remaining 5 digits can be arranged in \(5!\) ways.
\[
6 \times 5! = 6 \times 120 = 720
\]
2. **Starting with 2**: Similarly,
\[
6 \times 5! = 720
\]
3. **Starting with 3**:
\[
6 \times 5! = 720
\]
4. **Starting with 4**:
\[
6 \times 5! = 720
\]
5. **Starting with 5**: This cannot be considered as it would lead to numbers divisible by 5.
6. **Starting with 6**:
\[
6 \times 5! = 720
\]
7. **Starting with 7**:
\[
6 \times 5! = 720
\]
### Step 3: Cumulative counts
Now we can add these counts to find out where the 2004th number falls:
- Numbers starting with 1: 720 (1 to 720)
- Numbers starting with 2: 720 (721 to 1440)
- Numbers starting with 3: 720 (1441 to 2160)
- Numbers starting with 4: 720 (2161 to 2880)
Since 2004 falls between 1441 and 2160, we know the 2004th number starts with 3.
### Step 4: Fix the first digit and repeat the process
Now we fix the first digit as 3 and repeat the process for the remaining digits (1, 2, 4, 5, 6, 7) with the last digit not being 5.
1. **Starting with 31**:
\[
5 \times 4! = 5 \times 24 = 120
\]
(Numbers from 1441 to 1560)
2. **Starting with 32**:
\[
5 \times 4! = 120
\]
(Numbers from 1561 to 1680)
3. **Starting with 34**:
\[
5 \times 4! = 120
\]
(Numbers from 1681 to 1800)
4. **Starting with 35**:
\[
5 \times 4! = 120
\]
(Numbers from 1801 to 1920)
5. **Starting with 36**:
\[
5 \times 4! = 120
\]
(Numbers from 1921 to 2040)
### Step 5: Fix the second digit and repeat
Since 2004 falls between 1921 and 2040, we know the 2004th number starts with 36.
Now we fix the first two digits as 36 and repeat the process for the remaining digits (1, 2, 4, 5, 7).
1. **Starting with 361**:
\[
4 \times 3! = 4 \times 6 = 24
\]
(Numbers from 1921 to 1944)
2. **Starting with 362**:
\[
4 \times 3! = 24
\]
(Numbers from 1945 to 1968)
3. **Starting with 364**:
\[
4 \times 3! = 24
\]
(Numbers from 1969 to 1992)
4. **Starting with 365**:
\[
4 \times 3! = 24
\]
(Numbers from 1993 to 2016)
Since 2004 falls between 1993 and 2016, we know the 2004th number starts with 365.
### Step 6: Fix the third digit and repeat
Now we fix the first three digits as 365 and repeat the process for the remaining digits (1, 2, 4, 7).
1. **Starting with 3651**:
\[
3 \times 2! = 3 \times 2 = 6
\]
(Numbers from 1993 to 1999)
2. **Starting with 3652**:
\[
3 \times 2! = 6
\]
(Numbers from 2000 to 2006)
Since 2004 falls between 2000 and 2006, we know the 2004th number starts with 3652.
### Step 7: Fix the fourth digit
Now we fix the first four digits as 3652 and repeat the process for the remaining digits (1, 4, 7).
1. **Starting with 36521**:
\[
2! = 2
\]
(Numbers from 2000 to 2001)
2. **Starting with 36524**:
\[
2! = 2
\]
(Numbers from 2002 to 2003)
Since 2004 falls between 2002 and 2003, we know the 2004th number starts with 36524.
### Step 8: Final digits
The last remaining digits are 1 and 7. The 2004th number is:
- 3652417 (the first arrangement of the remaining digits).
Thus, the 2004th number in the list is **3652417**.
